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It seems that there exists a number of important results that are very useful when dealing with Hausdorff spaces, but that are not true in general. I was wondering if we can provide a list of such facts (with proofs) in this page, so that it can be used for future references?

To be more specific, what are some relations that hold when we are dealing with Hausdorff spaces, and that are very useful in measure theory?

For example, in class when proving Lusin's theorem, we used some statements, like a compact set contains all its points of closure, the intersection of any number of compact sets is compact etc, which left some students confused.

Arbër Ll
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    https://en.wikipedia.org/wiki/Hausdorff_space#Equivalences and https://en.wikipedia.org/wiki/Hausdorff_space#Properties. – lhf Dec 11 '15 at 09:44
  • There's infinite number of statements that are not generally true, but true for Hausdorff spaces. Probably a lot of these are "important". – skyking Dec 11 '15 at 09:46
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    For an interesting topology that is not Hausdorff, see the Zariski topology and http://math.stackexchange.com/questions/161884/why-zariski-topology. – lhf Dec 11 '15 at 10:00

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Fact: If $X$ is a Hausdoff space, then any compact set $S\subset X$ is closed.

Proof: if $x\notin S$, then for each $s\in S$ we can find opens $U_s, V_s$ such that $x\in U_s$, $s\in V_s$, and $U_s \cap V_s = \emptyset$. Since $S$ is compact, $S$ is covered by some finite subcollection $V_{s_1}, \ldots , V_{s_n}$. It follows that $U_{s_1} \cap \cdots \cap U_{s_n}$ is an open neighborhood of $x$ disjoint from $S$.

Conversely, if all compact sets are closed in $X$, then points are closed, so $X$ is $T_1$. It follows that any topological space that isn't $T_1$ has non-closed compact sets.

Fact: A compact Hausdorff space is regular.

Proof: Suppose $x\notin C$, where $C$ a closed set. Since a closed set in a compact space is compact, we can proceed exactly as in the preceding proof to construct disjoint open sets $U,V$ with $x\in U$ and $C\subset V$.

In fact, every compact Hausdorff space is normal, which follows from iterating the above argument. But there are lots of non-regular compact spaces, like the two-point Sierpiński space. And gluing together two copies along the open point yields a three-point space that is neither regular nor normal.

Call a space $X$ "locally compact(1)" if every point has a compact neighborhood. Call a space $X$ "locally compact(2)" if every open neighborhood of a point contains a compact neighborhood.

Fact: If a Hausdorff space is locally compact(1), it is locally compact(2).

Proof: Without loss of generality, suppose that $X$ is compact, and take an open neighborhood $U$ of $x$. Since a compact Hausdorff space is regular, there is a closed neighborhood of $x$ contained in $U$. Since a closed set in a compact space is compact, the conclusion follows.

On the other hand, there are spaces that are even locally Hausdorff that are locally compact(1) but not locally compact(2), see here.

Fact: In a Hausdorff space, the intersection of two compact sets is compact.

Proof: Since any compact set is closed and any closed subset of a compact is compact, this is trivial.

There are lots of counterexamples for non-Hausdorff spaces. For example, we can take any compact connected Hausdorff space $X$ with at least two points, and let $U\subset X$ be a nontrivial open. Gluing two copies of $X$ along $U$ gives a space with an intersection of two compact sets homeomorphic to $U$. But $U$ is a non-closed subset of a Hasudorff space, so cannot be compact.

For example, if we glue two copies of $[0,1]$ along $(0,1]$ we get an interval with a doubled endpoint, a compact locally Hausdorff space, but such that the intersection $(0,1]$ of the two intervals is non-compact.

Community
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Andrew Dudzik
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Let $\{x_n\}$ be a convergent sequence in a Hausdorff space $X$. Then the limit point $x_n \to x$ is unique.

If not, let $x_n \to x_0$ and $x_1$. Since $X$ is Hausdorff we can separate $x_0$ and $x_1$ by open disjoint sets $U$ and $V$ respectively. By definition of convergence, we can find a large enough $N$ such that $x_n \in U$ and $x_n \in V$ for $n > N$; clearly a contradiction.

Lionel Ricci
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    The proof is not complete as it doesn't show that this isn't generally valid. – skyking Dec 11 '15 at 09:48
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    @skyking The proof is perfectly fine. – Najib Idrissi Dec 11 '15 at 10:01
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    @NajibIdrissi No, it doesn't prove that it isn't generally valid. The question asked for statements that are true for Hausdorff space, but not valid in general. A proof for such a statement would mean that it would need to prove that it is valid for any Hausdorff space, but also prove that it isn't generally valid. – skyking Dec 11 '15 at 10:34
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    @skyking: Lionel has given a perfectly valid proof of his statement. Completely answering the OP's question is then an utter triviality : every sequence in an indiscrete space having at least two points (there are at least $2^{\aleph_0}$ such sequences) converges to any point of that space and thus has non unique limit. – Georges Elencwajg Dec 11 '15 at 18:14
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Every nontrivial connected compact Hausdorff space is the union of two nontrivial disjoint connected sets. For example, $[0,2]=[0,1]\cup (0,2]$.

The one-point compactification of $\mathbb Q$ is a nontrivial connected compact $T_1$ space without this property.

Actually a lot of basic continuum theory goes out the window if we don't assume all spaces are Hausdorff.

Forever Mozart
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